Functions0%

Algebra · Topic 3 of 12

Functions

Video lesson5 worked examples

Theory

A function is a mathematical relationship often denoted using functional notation like f(x)f(x) or g(x)g(x).

You must be able to substitute values into a function to evaluate it, and determine a missing variable when a function's overall output is given.

The Golden Rule: f(something)f(\text{something}) means “replace every xx in the formula with that something”. Put the value in brackets when you substitute — it keeps signs and powers correct.

⚠️ Common Examiner Traps

  • Squaring a negative: f(3)f(-3) with an x2x^2 term gives (3)2=9(-3)^2 = 9, positive — not 9-9. Always bracket the substituted value.
  • f(x)=0f(x) = 0 is not the same as f(0)f(0): the first solves an equation for xx; the second substitutes x=0x = 0.
  • Reading the notation backwards: in g(a)=35g(a) = 35, the 3535 is the output — set the formula equal to it and solve for aa.

Worked examples

Example 1

Evaluating a function

A function is defined by f(x)=x25xf(x) = x^2 - 5x. Calculate f(3)f(-3).

Step 1: Substitute −3 in place of x: (3)25(3)(-3)^2 - 5(-3).

Step 2: Calculate: 9+159 + 15.

Answer: 24.

Example 2

Finding an unknown

A function is defined as g(t)=4t+7g(t) = 4t + 7. Given that g(a)=35g(a) = 35, find a.

Step 1: Set the expression equal to 35: 4a+7=354a + 7 = 35.

Step 2: Solve for a: 4a=28a=74a = 28 \rightarrow a = 7.

Example 3

Finding an unknown (fractional function)

A function is defined by f(x)=2+6xf(x) = 2 + \dfrac{6}{x}. Given that f(a)=5f(a) = 5, find aa.

Step 1: Substitute and set equal to 5: 2+6a=52 + \dfrac{6}{a} = 5.

Step 2: Subtract 2 from both sides: 6a=3\dfrac{6}{a} = 3.

Step 3: Multiply both sides by aa, then divide by 3: 6=3aa=26 = 3a \rightarrow a = 2.

Answer: a=2a = 2.

Example 4

Evaluating a non-polynomial function

A function is defined by f(x)=10sinxf(x) = 10\sin x^\circ. Evaluate f(30)f(30).

Step 1: Function notation works the same whatever the formula — substitute 3030 for xx: f(30)=10sin30f(30) = 10\sin 30^\circ.

Step 2: Use the exact value sin30=12\sin 30^\circ = \tfrac{1}{2}: 10×1210 \times \tfrac{1}{2}.

Answer: 5.

Example 5

🔗 Bringing it together

A function is defined by f(x)=x25xf(x) = x^2 - 5x. Find the values of xx for which f(x)=0f(x) = 0.

Step 1: f(x)=0f(x) = 0 means set the formula equal to zero — an equation to solve, not a substitution: x25x=0x^2 - 5x = 0.

Step 2: Factorise (common factor of xx): x(x5)=0x(x - 5) = 0.

Step 3: A product is zero when either factor is zero, so x=0x = 0 or x5=0x - 5 = 0.

Answer: x=0x = 0 or x=5x = 5.